New RF Signal Generator

RIGOL Technologies recently expanded its portfolio of RF Test solutions with the launch of the DSG800 Series RF Signal Generator. The series—which is targeted at engineers implementing Bluetooth, Wi-Fi, and other RF interfaces in embedded systems—covers output frequencies from 9 kHz to 3 GHz. It provides maximum output power up to 20 dBm and low SSB phase noise of –105 dBc/Hz, amplitude accuracy of ±0.5 dB, and frequency resolution 0.01 Hz at any frequency. An oven-controlled crystal oscillator timebase provides less than 5 ppb temperature stability and less than 30 ppb/year aging stability.

The DSG800 RF signal generators provide:

• Conventional sweep functions (step, list, logarithmic, and linear)
• Analog modulation functions including amplitude modulation (AM), frequency modulation (FM), phase modulation (ΦM), and pulse modulation
• An optional pulse train generation capability for translating serial data onto an RF link

There are two models in the DSG800 series. THe DSG815 (9 kHz to 1.5 GHz) costs \$1,999. The DSG830 (9 kHz to 3 GHz) costs \$3,599.

Source: RIGOL Technologies

Question 1—A divider is a logic module that takes two binary numbers and produces their numerical quotient (and optionally, the remainder). The basic structure is a series of subtractions and multiplexers, where the multiplexer uses the result of the subtraciton to select the value that gets passed to the next step. The quotient is formed from the bits used to control the multiplexers, and the remainder is the result of the last subtraction.

If it is implemented purely combinatorially, then the critical path through all of this logic is quite long (even with carry-lookahead in the subtractors) and the clock cycle must be very slow. What could be done to shorten the clock period without losing the ability to get a result on every clock?

Answer 1—Pretty much any large chunk of combinatorial logic can be pipelined in order to reduce the clock period. This allows it to produce more results in a given amount of time, at the expense of increasing the latency for any particular result.

Divider logic is very easy to pipeline, and the number of pipeline stages you can use is fairly arbitrary. You could insert a pipeline register after each subtract-mux pair, or you might choose to do two or more subtract-mux stages per pipeline register You could even go so far as to pipeline the subtracts and the muxes separately (or even pipeline *within* each subtract) in order to get the fastest possible clock speed, but this would be rather extreme.

The more pipeline registers you use, the shorter the critical path (and the clock period) can be, but you use more resources (the registers). Also, the overall latency goes up, since you need to account for the setup and propagation times of the pipeline registers in the clock period (in addition to the subtract-mux logic delays). This gets multiplied by the number of pipeline stages in order to compute the total latency.

Answer 2—If you don’t need the level of performance provided by a pipelined divider, you can computes the quotient serially, one bit at a time. You would just need one subtractor and one multiplexer, along with registers to hold the input values, quotient bits and the intermediate result.

You could potentially compute more than one bit per clock period using additional subtract-mux stages. This gives you the flexibility to trade off space and time as needed for a particular application.

Question 3—An engineer wanted to build an 8-MHz filter that had a very narrow bandwidth, so he used a crystal lattice filter like this:

However, when he built and tested his filter, he discovered that while it worked fine around 8 MHz, the attenuation at very high frequencies (e.g., >80 MHz) was very much reduced. What caused this?

Answer 3—The equivalent circuit for a quartz crystal is something like this:

The components across the bottom represent the mechanical resonance of the crystal itself, while the capacitor at the top represents the capacitance of the electrodes and holder. Typical values are:

• Cser: 10s of fF (yes, femtofarads, 10-15F)
• L: 10s of mH
• R: 10s of ohms
• Cpar: 10s of pF

The crystal has a series-resonant frequency based on just Cser and L. It has a relatively low impedance (basically just R) at this frequency.

It also has a parallel-resonant (sometimes called “antiresonant”) frequency when you consider the entire loop, including Cpar. Since Cser and Cpar are essentially in series, together they have a slightly lower capacitance than Cser alone, so the parallel-resonant frequency is slightly higher. The crystal’s impedance is very high at this frequency.

But at frequencies much higher than either of the resonant frequencies, you can see that the impedance of Cparalone dominates, and this just keeps decreasing with increasing frequency. This reduces the crystal lattice filter to a simple capacitive divider, which passes high freqeuncies with little attenuation.

Finally, calculate the value of Cpar required to give that value of capacitance when in series with Cser:

Note that all three equations can be combined into one, and this reduces to:

Build an Adequate Test Bench (EE Tip #127)

It’s in our makeup as engineers that we want to test our newly received boards as soon as possible. We just can’t wait to connect them to a power supply and then use our test bench equipment (e.g., generators, oscilloscopes, switches or LEDs, and so on) for simulation.

Circuit Cellar columnist Robert Lacoste’s clean, orderly workspace in Chaville, France.

But due to our haste, the result is usually a PCB under test lying on a crowded workbench in the middle of a mesh of test cables, alligator clamps, prototyping boards, and other probes. Experience shows that the probability of a short circuit or mismatched connection is high during this phase of engineering excitement.

Rather than requiring a mesh of test wires, it is often wise to develop a small test PCB that will drastically simplify the test phase. Here the ancillary board provided a clean way to connect a Microchip Technology ICD3 debugger, a JTAG emulator, a debug analyzer, and a power supply input.

Take your time: prepare a real test bench to which you can connect your board. It could be as simple as a clean desk with properly labeled wires, but you might also need to anticipate the design of a test PCB in order to simplify the cabling.—Robert Lacoste, “Mixed-Signal Designs,” CC25:25th Anniversary Issue, 2013.

PROBLEM 1
Can you name all of the signals in the original 25-pin RS-232 connector?

Pins 9, 10, 11, 18, and 25 are unassigned/reserved. The rest are:

 Pin Abbreviation Source Description 1 PG – Protective ground 2 TD DTE Transmitted data 3 RD DCE Received data 4 RTS DTE Request to send 5 CTS DCE Clear to send 6 DSR DCE Data Set Ready 7 SG – Signal ground 8 CD DCE Carrier detect 12 SCD DCE Secondary carrier detect 13 SCTS DCE Secondary clear to send 14 STD DTE Secondary transmitted data 15 TC DCE Transmitter clock 16 SRD DCE Secondary received data 17 RC DCE Receiver clock 19 SRTS DTE Secondary request to send 20 DTR DTE Data terminal ready 21 SQ DCE Signal quality 22 RI DCE Ring indicator 23 – DTE Data rate selector 24 ETC DTE External transmitter clock

PROBLEM 2
What is the key difference between a Moore state machine and a Mealy state machine?

The key difference between Moore and Mealy is that in a Moore state machine, the outputs depend only on the current state, while in a Mealy state machine, the outputs can also be affected directly by the inputs.

PROBLEM 3
What are some practical reasons you might choose one state machine over the other?

In practice, the difference between Moore and Mealy in most situations is not very important. However, when you’re trying to optimize the design in certain ways, it sometimes is.

Generally speaking, a Mealy machine can have fewer state variables than the corresponding Moore machine, which will save physical resources on a chip. This can be important in low-power designs.

On the other hand, a Moore machine will typically have shorter logic paths between flip-flops (total combinatorial gate delays), which will enable it to run at a higher clock speed than the corresponding Mealy machine.

PROBLEM 4
What is the key feature that distinguishes a DSP from any other general-purpose CPU?

Usually, the key distinguishing feature of a DSP when compared with a general-purpose CPU is that the DSP can execute certain signal-processing operations with few, if any, CPU cycles wasted on instructions that do not compute results.

One of the most basic operations in many key DSP algorithms is the MAC (multiply-accumulate) operation, which is the fundamental step used in matrix dot and cross products, FIR and IIR filters, and fast Fourier transforms (FFTs). A DSP will typically have a register and/or memory organization and a data path that enables it to do at least 64 MAC operations (and often many more) on unique data pairs in a row without any clocks wasted on loop overhead or data movement. General-purpose CPUs do not generally have enough registers to accomplish this without using additional instructions to move data between registers and memory.

PROBLEM 1
Construct an electrical circuit to find the values of Xa, Xb, and Xc in this system of equations:

21Xa – 10Xb – 10Xc = 1
–10Xa + 22Xb – 10Xc = –2
–10Xa – 10Xb + 20Xc = 10

Your circuit should include only the following elements:

one 1-Ω resistor
one 2-Ω resistor
three 10-Ω resistors
three ideal constant voltage sources
three ideal ammeters

The circuit should be designed so that each ammeter displays one of the values Xa, Xb, or Xc. Given that the Xa, Xb, and Xc values represent currents, what kind of circuit analysis yields equations in this form?

You get equations in this form when you do mesh analysis of a circuit. Each equation represents the sum of the voltages around one loop in the mesh.

PROBLEM 2
What do the coefficients on the left side of the equations represent? What about the constants on the right side?

The coefficients on the left side of each equation represent resistances. Resistance multiplied by current (the unknown Xa, Xb, and Xc values) yields voltage.
The “bare” numbers on the right side of each equation represent voltages directly (i.e., independent voltage sources).

PROBLEM 3
What is the numerical solution for the equations?

To solve the equations directly, start by solving the third equation for Xc and substituting it into the other two equations:

Xc = 1/2 Xa + 1/2 Xb + 1/2

21Xa – 10Xb – 5Xa – 5Xb – 5 = 1
–10Xa + 22Xb – 5Xa – 5Xb – 5 = –2

16Xa – 15Xb = 6
–15Xa + 17Xb = 3

Solve for Xa by multiplying the first equation by 17 and the second equation by 15 and then adding them:

272Xa – 255Xb = 102
–225Xa + 255Xb = 45

47Xa = 147 → Xa = 147/47

Solve for Xb by multiplying the first equation by 15 and the second equation by 16 and then adding them:

240Xa – 225Xb = 90
–240Xa + 272Xb = 48

47Xb = 138 → Xb = 138/47

Finally, substitute those two results into the equation for Xc:

Xc = 147/94 + 138/94 + 47/94 = 332/94 = 166/47

PROBLEM 4
Finally, what is the actual circuit? Draw a diagram of the circuit and indicate the required value of each voltage source.

The circuit is a mesh comprising three loops, each with a voltage source. The common elements of the three loops are the three 10-Ω resistors, connected in a Y configuration (see the figure below).

The values of the voltage sources in each loop are given directly by the equations, as shown. To verify the numeric solution calculated previously, you can calculate all of the node voltages around the outer loop, plus the voltage at the center of the Y, and ensure they’re self-consistent.

We’ll start by naming Va as ground, or 0 V:

Vb = Va + 2 V = 2 V

Vc = Vb + 2 Ω × Xb = 2V + 2 Ω × 138/47 A = 370/47 V = 7.87234 V

Vd = Vc + 1 Ω × Xa = 370/47 V + 1 Ω × 147/47A = 517/47 V = 11.000 V

Ve = Vd – 1 V = 11.000 V – 1.000 V = 10.000 V

Va = Ve – 10 V = 0 V

which is where we started.

The center node, Vf, should be at the average of the three voltages Va, Vc, and Ve:

0 V + 370/47 V + 10 V/3 = 840/141 V = 5.95745 V

We should also be able to get this value by calculating the voltage drops across each of the three 10-Ω resistors:

Va + (Xc – Xb) × 10 Ω = 0 V + (166 – 138)/47A × 10 Ω = 280/47 V = 5.95745 V

Vc + (Xb – Xa) × 10 Ω = 370/47V + (138-147)/47A × 10 Ω = 280/47 V = 5.95745 V

Ve + (Xa – Xc) × 10 Ω = 10 V + (147-166)/47 A × 10 Ω = 280/47 V = 5.95745 V

PROBLEM 1
What is the key difference between the following two C functions?

```#define VOLTS_FULL_SCALE 5.000
#define KPA_PER_VOLT 100.0
#define KPA_THRESHOLD 200.0

*/
{
float pressure = voltage * KPA_PER_VOLT;

return pressure > KPA_THRESHOLD;
}

{
float voltage_threshold = KPA_THRESHOLD / KPA_PER_VOLT;
float adc_threshold = voltage_threshold / VOLTS_FULL_SCALE;

}```

The first function, test_pressure(), converts the ADC reading to engineering units before making the threshold comparison. This is a direct, obvious way to implement such a function.

The second function, test_pressure2(), converts the threshold value to an equivalent ADC reading, so that the two can be compared directly.

The key difference is in performance. The first function requires that arithmetic be done on each reading before making the comparison. However, the calculations in the second function can all be performed at compile time, which means that the only run-time operation is the comparison itself.

PROBLEM 2
How many NAND gates would it take to implement the following translation table? There are five inputs and eight outputs. You may consider an inverter to be a one-input NAND gate.

 Inputs Outputs A B C D E F G H I J K L M 1 1 1 1 1 0 0 0 0 1 1 1 1 0 1 1 1 1 0 0 0 0 0 0 1 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 1 1 1 1

First of all, note that there are really only four inputs and three unique outputs for this function, since input E is always 1 and outputs GHI are always 0. The only real outputs are F, plus the groups JK and LM.

Since the other 27 input combinations haven’t been specified, we can take the output values associated with all of them as “don’t care.”

The output F is simply the inversion of input C.

The output JK is high only when A is high or D is low.

The output LM is high except when B is low and C is high.

Therefore, the entire function can be realized with a total of five gates:

PROBLEM 3
Quick history quiz: Who were the three companies who collaborated to create the initial standard for the Ethernet LAN?

The original 10-Mbps Ethernet standard was jointly developed by Digital Equipment Corp. (DEC), Intel, and Xerox. It was released in November 1980, and was commonly referred to as “DIX Ethernet.”

PROBLEM 4
What was the name of the wireless network protocol on which Ethernet was based? Where was it developed?

The multiple access with collision detection protocol that Ethernet uses was based on a radio protocol developed at the University of Hawaii. It was known as the “ALOHA protocol.”

Problem 1—Tom, an FPGA designer, is helping out on a system that handles standard-definition digital video at 27 MHz and stores it into an SDRAM that runs at 200 MHz. He discovered the following logic in the FPGA (see Figure 1).

Let’s see if we can work out what it does. To start with, what is the output of the XOR gate in?

Answer 1—When the 27-MHz clock goes from low to high, the first flip-flop changes state. Let’s say that its output goes from low to high as well. Then, when the clock goes from high to low, the second flip-flop’s output will become the same as the first.

On the clock’s next rising edge, the first flip-flop will change again, this time from high to low. And on the next falling edge, the second one will follow suit.

Putting it another way, following each rising edge of the clock, the two flip-flops are different. Following each falling edge, they’re the same. Since we’re feeding them into an XOR gate, the gate’s output will be high following the clock’s rising edge and low following the falling edge. In other words, the XOR gate’s is a replica of the clock signal itself!

Problem 2—Why is this necessary?

Answer 2—In many FPGA architectures, clock signals are automatically assigned to special clock routing resources, which are different from—and kept separate from—the routing resources used for “ordinary” signals. The tools actually discourage (or even prevent) you from using a clock as an input to a gate or to any input of a flip-flop other than the clock input.

Therefore, when you need to pass a clock into another timing domain as a signal, it becomes necessary to generate an ordinary signal that is a replica of the clock. This is one way to accomplish that.

Problem 3—What is the AND gate’s output?

Answer 3—The three flip-flops in the 200-MHz domain have a delayed versions of the (replica) 27-MHz clock signal. The first two function as a conventional synchronizer to minimize the effects of metastability. The third one, along with the AND gate, functions as an edge detector, generating a one-clock pulse in the 200-MHz clock domain following each rising edge of the 27-MHz clock. This pulse might be used, for example, to initiate a write request in the SDRAM for each video data word.

Problem 4—Tom decided to verify the circuit’s operation in his logic simulator, but immediately ran into a problem. What was the problem and what could be added to the circuit to make simulation possible?

Answer 4—There is a subtle problem here for a simulator: All of the flip-flops start out in the “unknown” state. Feeding that back (inverted) to the first flip-flop leaves it in an unknown state. The entire simulation will never get out of the unknown state, even though we can reason that it doesn’t matter which actual state the first flip-flop starts out in. The XOR gate’s output will be known after one full clock cycle. To fix this, it is necessary to explicitly reset the first flip-flop at the beginning of the simulation, then the rest of the circuit will simulate normally.

Problem 1
Suppose you have an ordinary switch mode buck regulator. The input voltage is 100 V, the switch’s duty cycle is exactly 50%, and you measure the output voltage as 70 V. Is this converter operating in continuous conduction mode or discontinuous conduction mode? How can you tell?

If a switch mode buck converter is operating in continuous conduction mode, then the output voltage is the fraction of the input voltage as defined by the duty cycle. 100 V × 0.5 would equal 50 V. Therefore, this converter is operating in discontinuous conduction mode.

Note that continuous conduction mode includes the case in which synchronous (active) rectification is being used and the current through the coil is allowed to reverse direction when the output is lightly loaded. The output voltage in relation to the input voltage will still be defined by the switch duty cycle.

Therefore, we also know that the regulator in question is not using synchronous rectification, but rather is using a diode instead.

Problem 2
Since a diode can be placed in a High-Impedance state (reverse-biased) or a Low-Impedance state (forward-biased), they are sometimes used to switch AC signals, including audio and RF. What determines the magnitude of a signal that a diode can switch?

When diodes are used for signal switching, there are two considerations with regard to the magnitude of the signal relative to the DC control signal:

• In the Blocking state, the reverse bias voltage must be greater than the peak signal voltage to prevent signal leakage. Also, a high-bias voltage reduces the parasitic capacitance through the diode. PIN diodes are often used for RF switching because of their ultra-low capacitance.
• In the On state, the forward DC control current through diode must be greater than the peak AC signal current, and it must be large enough so that the current doesn’t approach the diode curve’s “knee” too closely, introducing distortion.

Obviously, the diode needs to be rated for both the peak reverse voltage and the peak forward current created by the combination of the control signal and the application signal.

Problem 3
What common function does the following truth table represent?

 A B C X Y Z 0 0 0 ? 0 0 0 0 0 1 ? 0 0 1 0 1 0 ? 0 1 0 0 1 1 ? 0 0 1 1 0 0 ? 1 0 0 1 0 1 ? 0 0 1 1 1 0 ? 0 1 0 1 1 1 ? 0 0 1

The truth table implements a form of priority encoder:

Z is set if C is set, otherwise
Y is set if B is set, otherwise
X is set if A is set

In other words, C has the highest priority and A has the lowest. However, unlike conventional priority encoders that produce a binary output, this one produces a “one hot” encoding.

Problem 4
Write the equations for the logic that would implement the table.

The logic is quite straightforward:

Z = C
Y = B & !C
X = A & !B & !C

The answers to the Circuit Cellar 274 Engineering Quotient are now available. The problems and answers are listed below.

Problem 1—What is wrong with the name “programmable unijunction transistor?”

Answer 1—Unlike the original unijunction transistor—which really does have just a single junction—the programmable unijunction transistor (PUT) is actually a four-layer device that has three junctions, much like a silicon-controlled rectifier (SCR).

Problem 2—Given a baseband channel with 3-kHz bandwidth and a 40-dB signal-to-noise ratio (SNR), what is the theoretical capacity of this channel in bits per second?

Answer 2—The impulse response of an ideal channel with exactly 3 kHz of bandwidth is a sinc (i.e., sin(x)/x) pulse in the time domain that has nulls that are 1/6,000 s apart. This means you could send a series of impulses through this channel at a 6,000 pulses per second rate. And, if you sampled at exactly the correct instants on the receiving end, you could recover the amplitudes of each of those pulses with no interference from the other pulses on either side of it.

However, a 40-dB signal-to-noise ratio implies that the noise power is 1/10,000 of the maximum signal power. In terms of distinguishing voltage or current levels, this means you can send at most sqrt(10,000) = 100 distinct levels through the channel before they start to overlap, making it impossible to separate one from another at the receiving end.

100 levels translates to log2100 = 6.64 binary bits of information. This means the total channel capacity is 3,9840 bits/s (i.e., 6,000 pulses/s × 6.64 bits/pulse).

This is the basis for the Shannon-Hartley channel capacity theorem.

Problem 3—In general, is it possible to determine whether a system is linear and time-invariant (LTI) by simply examining its input and output signals?

Answer 3—In general, given an input signal and an output signal, you might be able to definitively state that the system is not linear and time-invariant (LTI), but you’ll never be able to definitively state that it is, only that it might be.

The general technique is to use information in the input signal to see whether the output signal can be composed from the input features. Input signals (e.g., impulses and steps) are easist to analyze, but other signals can also be analyzed.

Problem 4—One particular system has this input signal:

Figure 1

The output is given by:

Figure 2

Is this system LTI?

Answer 4—In this example, the input is a rectangular pulse that can be analyzed as the superposition of two step functions that are separated in time, one positive-going and the other negative-going. This makes the analysis easy, since you can see the initial response to the first step function then determine whether the response following the second step is a linear combination of two copies of the first part of the response.

In this case, the response to the first step function at t = 0 is that the output starts rising linearly, also at t = 0. The second (negative) input step function occurs at t = 0.5, and if the system is LTI, you would expect the output to also change what it’s doing at that time. In fact, you would expect the output to level off at whatever value it had reached at that time, because the LTI response to the second step should be a negative-going linear ramp, which, when added to the original response, should cancel out.

However, this is not the output signal received, so this system is definitely not LTI.

The answers to the Circuit Cellar 272 Engineering Quotient are now available. The problems and answers are listed below.

Problem 1—Why does the power dissipation of a Darlington transistor tend to be higher than that of a single bipolar transistor in switching applications?

Answer 1—In switching applications, a single transistor can saturate, resulting in a low VCE of 0.3 to 0.4 V. However, in a Darlington pair, the output transistor is prevented from saturating by the negative feedback provided by the driver transistor. If the collector voltage drops below the sum of the VBE of the output transistor (about 0.7 V) and the VCE(sat) of the driver transistor (about 0.3 V), the drive current to the output transistor is reduced, preventing it from going into saturation itself. Therefore, the effective VCE(sat) of the Darlington pair is 1 V or more, resulting in much higher dissipation at a given current level.

Problem 2—Suppose you have some 3-bit data, say, grayscale values for which 000 = black and 111 = white. You have a display device that takes 8-bit data, and you want to extend the bit width of your data to match.

If you just pad the data with zeros, you get the value 11100000 for white, which is not full white for the 8-bit display—that would be 11111111. What can you do?

Answer 2—One clever trick is to repeat the bits you have as many times as necessary to fill the output field width. For example, if the 3-bit input value is ABC, the output value would be ABCABCAB. This produces the following mapping, which interpolates nicely between full black and full white (see Table 1). Note that this mapping preserves the original bits; if you want to go back to the 3-bit representation, just take the MSBs and you have the original data.

 3-bit INPUT 8-bit OUTPUT 000 00000000 001 00100100 010 01001001 011 01101101 100 10010010 101 10110110 110 11011011 111 11111111

Problem 3—Can an induction motor (e.g., squirrel-cage type) be used as a generator?

Answer 3—Believe it or not, yes it can.

An induction motor has no electrical connections to the rotor; instead, a magnetic field is induced into the rotor by the stator. The motor runs slightly slower than “synchronous” speed—typically 1725 or 3450 rpm when on 60 Hz power.

If the motor is provided with a capacitive load, is driven at slightly higher than synchronous speed (1875 or 3750 rpm), and has enough residual magnetism in the rotor to get itself going, it will generate power up to approximately its rating as a motor. The reactive current of the load capacitor keeps the rotor energized in much the same way as when it is operating as a motor.

Problem 4—In Figure 1, why does this reconstruction of a 20-kHz sinewave sampled at 44.1 kHz show ripple in its amplitude?

Answer 4—The actual sampled data, represented by the square dots in the diagram, contains equal levels of Fsignal (the sine wave) and Fsample-Fsignal (one of the aliases of the sinewave). Any reconstruction filter is going to have difficulty passing the one and eliminating the other, so you inevitably get some of the alias signal, which, when added to the desired signal, produces the “modulation” you see.

In the case of a software display of a waveform on a computer screen (e.g., such as you might see in software used to edit audio recordings), they’re probably using an FIR low-pass filter (sin(x)/x coefficients) windowed to some finite length. A shorter window gives faster drawing times, so they’re making a tradeoff between visual fidelity and interactive performance. The windowing makes the filter somewhat less than brick-wall, so you get the leakage of the alias and the modulation.

In the case of a real audio D/A converter, even with oversampling you can’t get perfect stopband attenuation (and you must always do at least some of the filtering in the analog domain), so once again you see the leakage and modulation.

In this example, Fsignal = 0.9×Fnyquist, so Falias = 1.1×Fnyquist and Falias/Fsignal = 1.22. To eliminate the visible artifacts, the reconstruction filter would need to have a slope of about 60dB over this frequency span, or about 200 dB/octave.

The answers to the Circuit Cellar 270 Engineering Quotient are now available. The problems and answers are listed below.

Problem 1: Given a microprocessor that has hardware support for just one level of priority for interrupts, is it possible to implement multiple priorities in software? If so, what are the prerequisites that are required?

Answer 1: Yes, given a few basic capabilities, it is possible to implement multiple levels of interrupt priority in software. The basic requirements are that it must be possible to reenable interrupts from within an interrupt service routine (ISR) and that the different interrupt sources can be individually masked.

Question 2: What is the basic scheme for implementing software interrupt priorities?

Answer 2: In normal operation, all the interrupt sources are enabled, along with the processor’s global-interrupt mask.

When an interrupt occurs, the global interrupt mask is disabled and the “master” ISR is entered. This code must (quickly) determine which interrupt occurred, disable that interrupt and all lower-priority interrupts at their sources, then reenable the global-interrupt mask before jumping to the ISR for that interrupt. This can often be facilitated by precomputing a table of interrupt masks for each priority level.

Question 3: What are some of the problems associated with software interrupt priorities?

Answer 3: For one thing, the start-up latency of all the ISRs is increased by the time spent in the “master” ISR. This can be a problem in time-critical systems. This scheme enables interrupts to be nested, so the stack must be large enough to handle the worst-case nesting of ISRs, on top of the worst-case nesting of non-interrupt subroutine calls.

Finally, it is very tricky to do this in anything other than Assembly language. If you want to use a high-level language, you’ll need to be intimately familiar with the language’s run-time library and how it handles interrupts and reentrancy, in general.

Answer 4: Yes, on most such processors, you can execute a subroutine call to a “return from interrupt” instruction while still in the master ISR, which will then return to the master ISR, but with interrupts enabled.

Check to see whether the “return from interrupt” affects any other processor state (e.g., popping a status word from the stack) and prepare the stack accordingly.

Also, beware that another interrupt could occur immediately thereafter, and make sure the master ISR is reentrant beyond that point.

Contributed by David Tweed

Electrical Engineering Tools & Preparation (CC 25th Anniversary Issue Preview)

Electrical engineering is frequently about solving problems. Success requires a smart plan of action and the proper tools. But as all designers know, getting started can be difficult. We’re here to help.

You don’t have to procrastinate or spend a fortune on tools to start building your own electronic circuits. As engineer/columnist Jeff Bachiochi has proved countless times during the past 25 years,  there are hardware and software tools that fit any budget. In Circuit Cellar‘s 25th Anniversary issue, he offers some handy tips on building a tool set for successful electrical engineering. Bachiochi writes:

In this essay, I’ll cover the “build” portion of the design process. For instance, I’ll detail various tips for prototyping, circuit wiring, enclosure preparation, and more. I’ll also describe several of the most useful parts and tools (e.g., protoboards, scopes, and design software) for working on successful electronic design projects. When you’re finished with this essay, you’ll be well on your way to completing a successful electronic design project.

The Prototyping Process

Prototyping is an essential part of engineering. Whether you’re working on a complicated embedded system or a simple blinking LED project, building a prototype can save you a lot of time, money, and hassle in the long run. You can choose one of three basic styles of prototyping: solderless breadboard, perfboard, and manufactured PCB. Your project goals, your schedule, and your circuit’s complexity are variables that will influence your choice. (I am not including styles like flying leads and wire-wrapping.)

Prototyping Tools

The building phase of a design might include wiring up your circuit design and altering an enclosure to provide access to any I/O on the PCB. Let’s begin with some tools that you will need for circuit prototyping.

The nearby photo shows a variety of small tools that I use when wiring a perfboard or assembling a manufactured PCB. The needle-nose pliers/cutter is the most useful.

These are my smallest hand tools. With them I can poke, pinch, bend, cut, smooth, clean, and trim parts, boards, and enclosures. I can use the set of special driver tips to open almost any product that uses security screws.

Don’t skimp on this; a good pair will last many years. …

Once everything seems to be in order, you can fill up the sockets. You might need to provide some stimulus if you are building something like a filter. A small waveform generator is great for this. There are even a few hand probes that will provide outputs that can stimulate your circuitry. An oscilloscope might be the first “big ticket” item in which you invest. There are some inexpensive digital scope front ends that use an app running on a PC for display and control, but I suggest a basic analog scope (20 MHz) if you can swing it (starting at less than \$500).

If the circuit doesn’t perform the expected task, you should give the wiring job a quick once over. Look to see if something is missing, such as an unconnected or misconnected wire. If you don’t find something obvious, perform a complete continuity check of all the components and their connections using an ohmmeter.

I use a few different meters. One has a transistor checker. Another has a high-current probe. For years I used a small battery-powered hand drill before purchasing the Dremel and drill press. The tweezers are actually an SMT parts measurer. Many are unmarked and impossible to identify without using this device (and the magnifier).

It usually will be a stupid mistake. To do a complete troubleshooting job, you’ll need to know how the circuit is supposed to work. Without that knowledge, you can’t be expected to know where to look and what to look for.

Make a Label

You’ll likely want to label your design… Once printed, you can protect a label by carefully covering it with a single strip of packing tape.

The label for this project came straight off a printer. Using circuit-mount parts made assembling the design a breeze.

A more expensive alternative is to use a laminating machine that puts your label between two thin plastic sheets. There are a number of ways to attach your label to an enclosure. Double-sided tape and spray adhesive (available at craft stores) are viable options.”

Check out the upcoming anniversary issue for Bachiochi’s complete essay.

Debugging USB Firmware

You’ve written firmware for your USB device and are ready to test it. You attach the device to a PC and the hardware wizard announces: “The device didn’t start.” Or, the device installs but doesn’t send or receive data. Or, data is being dropped, the throughput is low, or some other problem presents itself. What do you do?

This article explores tools and techniques to debug the USB devices you design. The focus is on USB 2.0 devices, but much of the information also applies to developing USB 3.0 (SuperSpeed) devices and USB hosts for embedded systems.

VIEWING BUS TRAFFIC

If you do anything beyond a small amount of USB developing, a USB protocol analyzer will save you time and trouble. Analyzers cost less than they used to and are well worth the investment.

A hardware-based analyzer connects in a cable segment upstream from the device under test (see Photo 1).

Photo 1: The device under test connects to the analyzer, which
captures the data and passes it unchanged to the device’s host. The
cable on the back of the analyzer carries the captured data to the
analyzer’s host PC for display.

You can view the data down to each packet’s individual bytes and see exactly what the host and device did and didn’t send (see Photo 2).

Photo 2: This bus capture shows the host’s request for a configuration
descriptor and the bytes the device sent in response. Because the endpoint’s
maximum packet size is eight, the device sends the first 8 bytes in one
transaction and the final byte in a second transaction.

An analyzer can also decode data to show standard USB requests and class-specific data (see Photo 3).

Photo 3: This display decodes a received configuration descriptor and its subordinate descriptors.

To avoid corrupted data caused by the electrical effects of the analyzer’s connectors and circuits, use short cables (e.g., 3’ or less) to connect the analyzer to the device under test.

Software-only protocol analyzers, which run entirely on the device’s host PC, can also be useful. But, this kind of analyzer only shows data at the host-driver level, not the complete packets on the bus.

DEVELOPMENT STRATEGIES

The first rule for developing USB device firmware is to remember that the host computer controls the bus. Devices just need to respond to received data and events. Device firmware shouldn’t make assumptions about what the host will do next.

For example, some flash drives work under Windows but break when attached to a host with an OS that sends different USB requests or mass-storage commands, sends commands in a different order, or detects errors Windows ignores. This problem is so common that Linux has a file, unusual_devs.h, with fixes for dozens of misbehaving drives.

The first line of defense in writing USB firmware is the free USB-IF Test Suite from the USB Implementers Forum (USB-IF), the trade group that publishes the USB specifications. During testing, the suite replaces the host’s USB driver with a special test driver. The suite’s USB Command Verifier tool checks for errors (e.g., malformed descriptors, invalid responses to standard USB requests, responses to Suspend and Resume signaling, etc.). The suite also provides tests for devices in some USB classes, such as human interface devices (HID), mass storage, and video.

Running the tests will usually reveal issues that need attention. Passing the tests is a requirement for the right to display the USB-IF’s Certified USB logo.

LAYERED COMMUNICATIONS

Like networks, USB communications have layers that isolate different logical functions (see Table 1).

Table 1: USB communications use layers, which are each responsible for a
specific logical function.

The USB protocol layer manages USB transactions, which carry data packets to and from device endpoints. A device endpoint is a buffer that is a source or sink of data at the device. The host sends data to Out endpoints and receives data from In endpoints. (Even though endpoints are on devices, In and Out are defined from the host’s perspective.)

The device layer manages USB transfers, with each transfer moving a chunk of data consisting of one or more transactions. To meet the needs of different peripherals, the USB 2.0 specification defines four transfer types: control, interrupt, bulk, and isochronous.

The function layer manages protocols specific to a device’s function (e.g., mouse, printer, or drive). The function protocols may be a combination of USB class, industry, and vendor-defined protocols.

CONTROLLER ARCHITECTURES

The layers supported by device firmware vary with the device hardware. At one end of the spectrum, a Future Technology Devices International (FTDI) FT232R USB UART controller handles all the USB protocols in hardware. The chip has a USB device port that connects to a host computer and a UART port that connects to an asynchronous serial port on the device.

Device firmware reads and writes data on the serial port, and the FT232R converts it between the USB and UART protocols. The device firmware doesn’t have to know anything about USB. This feature has made the FT232R and similar chips popular!

An example of a chip that is more flexible but requires more firmware support is Microchip Technology’s PIC18F4550 microcontroller, which has an on-chip, full-speed USB device controller. In return for greater firmware complexity, the PIC18F4550 isn’t limited to a particular host driver and can support any USB class or function.

Each of the PIC18F4550’s USB endpoints has a series of registers—called a buffer descriptor table (BDT)—that store the endpoint buffer’s address, the number of bytes to send or receive, and the endpoint’s status. One of the BDT’s status bits determines the BDT’s ownership. When the CPU owns the BDT, firmware can write to the registers to prepare to send data or to retrieve received data. When the USB module owns the BDT, the endpoint can send or receive data on the bus.

To send a data packet from an In endpoint, firmware stores the bytes’ starting address to send and the number of bytes and sets a register bit to transfer ownership of the BDT to the USB module. The USB module sends the data in response to a received In token packet on the endpoint and returns BDT ownership to the CPU so firmware can set up the endpoint to send another packet.

To receive a packet on an Out endpoint, firmware stores the buffer’s starting address for received bytes and the maximum number of bytes to receive and transfers ownership of the BDT to the USB module. When data arrives, the USB module returns BDT ownership to the CPU so firmware can retrieve the data and transfer ownership of the BDT back to the USB module to enable the receipt of another packet.

Other USB controllers have different architectures and different ways of managing USB communications. Consult your controller chip’s datasheet and programming guide for details. Example code from the chip vendor or other sources can be helpful.

DEBUGGING TRANSACTIONS

A USB 2.0 transaction consists of a token packet and, as needed, a data packet and a handshake packet. The token packet identifies the packet’s type (e.g., In or Out), the destination device and endpoint, and the data packet direction.

The data packet, when present, contains data sent by the host or device. The handshake packet, when present, indicates the transaction’s success or failure.

The data and handshake packets must transmit quickly after the previous packet, with only a brief inter-packet delay and bus turnaround time, if needed. Thus, device hardware typically manages the receiving and sending of packets within a transaction.

For example, if an endpoint’s buffer has room to accept a data packet, the endpoint stores the received data and returns ACK in the handshake packet. Device firmware can then retrieve the data from the buffer. If the buffer is full because firmware didn’t retrieve previously received data, the endpoint returns NAK, requiring the host to try again. In a similar way, an In endpoint will NAK transactions until firmware has loaded the endpoint’s buffer with data to send.

Fine tuning the firmware to quickly write and retrieve data can improve data throughput by reducing or eliminating NAKs. Some device controllers support ping-pong buffers that enable an endpoint to store multiple packets, alternating between the buffers, as needed.

LOST DATA

In all but isochronous transfers, a data-toggle value in the data packet’s packet identification (PID) field guards against missed or duplicate data packets. If you’re debugging a device where data is transmitting on the bus and the receiver is returning ACK but ignoring or discarding the data, chances are good that the device isn’t sending or expecting the correct data-toggle value. Some device controllers handle the data toggles completely in hardware, while others require some firmware control.

Each endpoint maintains its own data toggle. The values are DATA0 (0011B) and DATA1 (1011B). Upon detecting an incoming data packet, the receiver compares its data toggle’s state with the received data toggle. If the values match, the receiver toggles its value and returns ACK, causing the sender to toggle its value for the next transaction.

The next received packet should contain the opposite data toggle, and again the receiver toggles its bit and returns ACK. Except for control transfers, the data toggle on each end continues to alternate in each transaction. (Control transfers always use DATA0 in the Setup stage, toggle the value for each transaction in the Data stage, and use DATA1 in the Status stage.)

If the receiver returns NAK or no response, the sender doesn’t toggle its bit and tries again with the same data and data toggle. If a receiver returns ACK, but for some reason the sender doesn’t see the ACK, the sender thinks the receiver didn’t receive the data and tries again using the same data and data toggle. In this case, the repeated data receiver ignores the data, doesn’t toggle the data toggle, and returns ACK, resynchronizing the data toggles. If the sender mistakenly sends two packets in a row with the same data-toggle value, upon receiving the second packet, the receiver ignores the data, doesn’t toggle its value, and returns ACK.

DEFINING A TRANSFER

All USB devices must support control transfers and may support other transfer types. Control transfers provide a structure for sending requests but have no guaranteed delivery time. Interrupt transfers have a guaranteed maximum latency (i.e., delay) between transactions, but the host permits less bandwidth for interrupt transfers compared to other transfer types. Bulk transfers are the fastest on an otherwise idle bus, but they have no guaranteed delivery time, and thus can be slow on a busy bus. Isochronous transfers have guaranteed delivery time but no built-in error correction.

A transfer’s amount of data depends in part on the higher-level protocol that determines the data packets’ contents. For example, a keyboard sends keystroke data in an interrupt transfer that consists of one transaction with 8 data bytes. To send a large file to a drive, the host typically uses one or more large transfers consisting of multiple transactions. For a high-speed drive, each transaction, except possibly the last one, has 512 data bytes, which is the maximum-allowed packet size for high-speed bulk endpoints.

What determines a transfer’s end varies with the USB class or vendor protocol. In many cases, a transfer ends with a short packet, which is a packet that contains less than the packet’s maximum-allowed data bytes. If the transfer has an even multiple of the packet’s maximum-allowed bytes, the sender may indicate the end of the transfer with a zero-length packet (ZLP), which is a data packet with a PID and error-checking bits but no data.

For example, USB virtual serial-port devices in the USB communications device class use short packets to indicate the transfer’s end. If a device has sent data that is an exact multiple of the endpoint’s maximum packet size and the host sends another In token packet, the endpoint should return a ZLP to indicate the data’s end.

DEBUGGING ENUMERATION

Upon device attachment, in a process called enumeration, the host learns about the device by requesting a series of data structures called descriptors. The host uses the descriptors’ information to assign a driver to the device.

If enumeration doesn’t complete, the device doesn’t have an assigned driver, and it can’t perform its function with the host. When Windows fails to find an appropriate driver, the setupapi.dev.log file in Windowsinf (for Windows 7) can offer clues about what went wrong. A protocol analyzer shows if the device returned all requested descriptors and reveals mistakes in the descriptors.

During device development, you may need to change the descriptors (e.g., add, remove, or edit an endpoint descriptor). Windows has the bad habit of remembering a device’s previous descriptors on the assumption that a device will never change its descriptors. To force Windows to use new descriptors, uninstall then physically remove and reattach the device from Windows Device Manager. Another option is to change the device descriptor’s product ID to make the device appear as a different device.

DEBUGGING TRANSFERS

Unlike the other transfer types, control transfers have multiple stages: setup, (optional) data, and status. Devices must accept all error-free data packets that follow a Setup token packet and return ACK. If the device is in the middle of another control transfer and the host sends a new Setup packet, the device must abandon the first transfer and begin the new one. The data packet in the Setup stage contains important information firmware should completely decode (see Table 2).

Table 2: Device firmware should fully decode the data received in a control transfer’s Setup stage. (Source: USB Implementers Forum, Inc.)

The wLength field specifies how many bytes the host wants to receive. A device shouldn’t assume how much data the host wants but should check wLength and send no more than the requested number of bytes.

For example, a request for a configuration descriptor is actually a request for the configuration descriptor and all of its subordinate descriptors. But, in the first request for a device’s configuration descriptor, the host typically sets the wLength field to 9 to request only the configuration descriptor. The descriptor contains a wTotalLength value that holds the number of bytes in the configuration descriptor and its subordinate descriptors. The host then resends the request setting wLength to wTotalLength or a larger value (e.g., FFh). The device returns the requested descriptor set up to wTotalLength. (Don’t assume the host will do it this way. Always check wLength!)

Each Setup packet also has a bmRequestType field. This field specifies the data transfer direction (if any), whether the recipient is the device or an interface or endpoint, and whether the request is a standard USB request, a USB class request, or a vendor-defined request. Firmware should completely decode this field to correctly identify received requests.

A composite device has multiple interfaces that function independently. For example, a printer might have a printer interface, a mass-storage interface for storing files, and a vendor-specific interface to support vendor-defined capabilities. For requests targeted to an interface, the wIndex field typically specifies which interface applies to the request.

INTERRUPT TRANSFER TIMING

For interrupt endpoints, the endpoint descriptor contains a bInterval value that specifies the endpoint’s maximum latency. This value is the longest delay a host should use between transaction attempts.

A host can use the bInterval delay time or a shorter period. For example, if a full-speed In endpoint has a bInterval value of 10, the host can poll the endpoint every 1 to 10 ms. Host controllers typically use predictable values, but a design shouldn’t rely on transactions occurring more frequently than the bInterval value.

Also, the host controller reserves bandwidth for interrupt endpoints, but the host can’t send data until a class or vendor driver provides something to send. When an application requests data to be sent or received, the transfer’s first transaction may be delayed due to passing the request to the driver and scheduling the transfer.

Once the host controller has scheduled the transfer, any additional transaction attempts within the transfer should occur on time, as defined by the endpoint’s maximum latency. For this reason, sending a large data block in a single transfer with multiple transactions can be more efficient than using multiple transfers with a portion of the data in each transfer.

DEVICE FUNCTIONS

Most devices’ functions fit a defined USB class (e.g., mass storage, printer, audio, etc.). The USB-IF’s class specifications define protocols for devices in the classes.

For example, devices in the HID class must send and receive all data in data structures called reports. The supported report’s length and the meaning of its data (e.g., keypresses, mouse movements, etc.) are defined in a class-specific report descriptor.

If your HID-class device is sending data but the host application isn’t seeing the data, verify the number of bytes the device is sending matches the number of bytes in a defined report. The device should prepend a report-ID byte to the data only if the HID supports report IDs other than the zero default value.

In many devices, class specifications define class-specific requests or other requirements. For example, a mass storage device that uses the bulk-only protocol must provide a unique serial number in a string descriptor. Carefully read and heed any class specifications that apply to your device!

Many devices also support industry protocols to perform higher-level functions. Printers typically support one or more printer-control languages (e.g., PCL and Postscript). Mass-storage devices support SCSI commands to transfer data blocks and a file system (e.g., FAT32) to define a directory structure.

The higher-level industry protocols don’t depend on a particular hardware interface, so there is little about debugging them that is USB-specific. But, because these protocols can be complicated, example code for your device can be helpful.

In the end, much about debugging USB firmware is like debugging any hardware or software. A good understanding of how the communications should work provides a head start on writing good firmware and finding the source of any problems that may appear.

Jan Axelson is the author of USB Embedded Hosts, USB Complete, and Serial Port Complete. Jan’s PORTS web forum is available at www.lvr.com.

RESOURCES

Jan Axelson’s Lakeview Research, “USB Development Tools: Protocol analyzers,” www.lvr.com/development_tools.htm#analyzers.

DIY Green Energy Design Projects

Ready to start a low-power or energy-monitoring microcontroller-based design project? You’re in luck. We’re featuring eight award-winning, green energy-related designs that will help get your creative juices flowing.

The projects listed below placed at the top of Renesas’s RL78 Green Energy Challenge.

Electrostatic Cleaning Robot: Solar tracking mirrors, called heliostats, are an integral part of Concentrating Solar Power (CSP) plants. They must be kept clean to help maximize the production of steam, which generates power. Using an RL78, the innovative Electrostatic Cleaning Robot provides a reliable cleaning solution that’s powered entirely by photovoltaic cells. The robot traverses the surface of the mirror and uses a high voltage AC electric field to sweep away dust and debris.

Parts and circuitry inside the robot cleaner

Cloud Electrofusion Machine: Using approximately 400 times less energy than commercial electrofusion machines, the Cloud Electrofusion Machine is designed for welding 0.5″ to 2″ polyethylene fittings. The RL78-controlled machine is designed to read a barcode on the fitting which determines fusion parameters and traceability. Along with the barcode data, the system logs GPS location to an SD card, if present, and transmits the data for each fusion to a cloud database for tracking purposes and quality control.

Inside the electrofusion machine (Source: M. Hamilton)

The Sun Chaser: A GPS Reference Station: The Sun Chaser is a well-designed, solar-based energy harvesting system that automatically recalculates the direction of a solar panel to ensure it is always facing the sun. Mounted on a rotating disc, the solar panel’s orientation is calculated using the registered GPS position. With an external compass, the internal accelerometer, a DC motor and stepper motor, you can determine the solar panel’s exact position. The system uses the Renesas RDKRL78G13 evaluation board running the Micrium µC/OS-III real-time kernel.

[Video: ]

Water Heater by Solar Concentration: This solar water heater is powered by the RL78 evaluation board and designed to deflect concentrated amounts of sunlight onto a water pipe for continual heating. The deflector, armed with a counterweight for easy tilting, automatically adjusts the angle of reflection for maximum solar energy using the lowest power consumption possible.

RL78-based solar water heater (Source: P. Berquin)

Air Quality Mapper: Want to make sure the air along your daily walking path is clean? The Air Quality Mapper is a portable device designed to track levels of CO2 and CO gasses for constructing “Smog Maps” to determine the healthiest routes. Constructed with an RDKRL78G13, the Mapper receives location data from its GPS module, takes readings of the CO2 and CO concentrations along a specific route and stores the data in an SD card. Using a PC, you can parse the SD card data, plot it, and upload it automatically to an online MySQL database that presents the data in a Google map.

Air quality mapper design (Source: R. Alvarez Torrico)

Wireless Remote Solar-Powered “Meteo Sensor”: You can easily measure meteorological parameters with the “Meteo Sensor.” The RL78 MCU-based design takes cyclical measurements of temperature, humidity, atmospheric pressure, and supply voltage, and shares them using digital radio transceivers. Receivers are configured for listening of incoming data on the same radio channel. It simplifies the way weather data is gathered and eases construction of local measurement networks while being optimized for low energy usage and long battery life.

The design takes cyclical measurements of temperature, humidity, atmospheric pressure, and supply voltage, and shares them using digital radio transceivers. (Source: G. Kaczmarek)

Portable Power Quality Meter: Monitoring electrical usage is becoming increasingly popular in modern homes. The Portable Power Quality Meter uses an RL78 MCU to read power factor, total harmonic distortion, line frequency, voltage, and electrical consumption information and stores the data for analysis.

The portable power quality meter uses an RL78 MCU to read power factor, total harmonic distortion, line frequency, voltage, and electrical consumption information and stores the data for analysis. (Source: A. Barbosa)

High-Altitude Low-Cost Experimental Glider (HALO): The “HALO” experimental glider project consists of three main parts. A weather balloon is the carrier section. A glider (the payload of the balloon) is the return section. A ground base section is used for communication and display telemetry data (not part of the contest project). Using the REFLEX flight simulator for testing, the glider has its own micro-GPS receiver, sensors and low-power MCU unit. It can take off, climb to pre-programmed altitude and return to a given coordinate.

High-altitude low-cost experimental glider (Source: J. Altenburg)

Problem 1: A transformer’s windings, when measured individually (all other windings disconnected), have a certain amount of inductance. If you have a 1:1 transformer (both windings have the same inductance) and connect the windings in series, what value of inductance do you get?

Answer 1: Assuming you connect the windings in-phase, you’ll have double the number of turns, so the resulting inductance will be about four times the inductance of one winding alone.

If you hook them up out of phase, the inductance will cancel out and you’ll be left with the resistance of the wire and a lot of parasitic inter-winding capacitance.

Problem 2: If you connect the windings in parallel, what value of inductance do you get?

Answer 2: With the two windings connected in-phase and in parallel, the inductance will be exactly the same as the single-winding case. But the resulting inductor will be able to handle twice the current, as long as the core itself doesn’t saturate.

Question 3: Suppose you have a 32-bit word in your microprocessor, and you want to count how many contiguous strings ones that appear in it. For example, the word “01110001000111101100011100011111” contains six such strings. Can you come up with an algorithm that uses simple shifts, bitwise logical and arithmetic operators, but —here’s the twist—does not require iterating over each bit in the word?

Answer 3: Here’s a solution that iterates over the number of strings, rather than the number of bits in the word.

```int nstrings (unsigned long int x)
{
int result = 0;

/* convert x into a word that has a '1' for every
* transition from 0 to 1 or 1 to 0 in the original
* word.
*/
x ^= (x << 1);

/* every pair of ones in the new word represents
* a string of ones in the original word. Remove
* them two at a time and keep count.
*/
while (x) {
/* remove the lowest set bit from x; this
* represents the start of a string of ones.
*/
x &= ~(x & -x);
++result;

/* remove the next set bit from x; this
* represents the end of that string of ones.
*/
x &= ~(x & -x);
}
return result;
}```

Problem 4: For the purpose of timing analysis, the operating conditions of an FPGA are sometimes known as “PVT,” which stands for “process, voltage, and temperature.” Voltage and temperature are pretty much self-explanatory, but what does process mean in this context?

Answer 4: The term process in this case refers to the manufacturing process at the plant where they make the FPGA. It’s a measure of the statistical variability of the physical characteristics from chip to chip as they come off the line.
This includes everything from mask alignment to etching times to doping levels. These things affect electrical parameters such as sheet and contact resistance, actual transistor gains, and thresholds and parasitic capacitances.
These kinds of variations are unavoidable, and the P in PVT is an attempt to account for their effects in the timing analysis. The idea is to make the analysis conservative enough so that your design will work reliably despite these variations.

Contributed by David Tweed